Let $a, b, c, d$ be points in the Euclidean plane. Suppose that $abcd$ is a non-degenerate rectangle, and that the length of the line segment $ab$ is at least as big as the length of the line segment $bc$. Denote by $P$ the perpendicular bisector of $ab$, and let $p \in P$ lie in the exterior of $abcd$. Can a circle be drawn through $p$ whose interior contains $abcd$?
Yes, it is possible to draw a circle through $p$, whose interior contains the closed (in the topological sense) rectangle $abcd$. I will give a justified construction. But first, I will list three facts that I will require for the construction's justification.
Preliminaries
Fact 1 (Scalene Inequality) For every $\triangle abc$, if $\angle b < \angle c$, then $\overline{ac} < \overline{ab}$. (I use the notation $\overline{xy}$ to denote the length of the line segment between $x$ and $y$.)
Proof [V] Theorem 4.3.1, p. 77. $\square$
Fact 2 (Triangle Inequality) If $a$, $b$, and $c$ are noncollinear, then $\overline{ab} + \overline{bc} > \overline{ac}$.
Proof [V] Theorem 4.3.2, p. 78. $\square$
Fact 3 The three perpendicular bisectors of the sides of a triangle are concurrent and meet at a point that is equidistant from all three vertices of the triangle.
Proof [V] Corollary 8.2.5, p. 203. $\square$
I will use other facts without proof if I consider them to be intuitive, for instance the fact that the interior of a circle is convex, which is a concise way of saying that whenever $x$ and $y$ are two points lying in the interior of a circle, the entire line segment between $x$ and $y$ lies in the interior of this circle.
Construction
Denote by $o$ the point of intersection of the perpendicular bisectors of the sides of $\triangle abp$ (note that $P$ is one of these bisectors). Then, by fact 3 there exists a circle, call it $r$, whose center is $o$, and such that $a, b, p \in r$.
Denote by $s$ the intersection of $ab$ and $P$. $o$ must lie on the ray $\overrightarrow{ps}$ from $p$ in the direction of $s$ (note that $\overrightarrow{ps}$ extends beyond $s$, to "infinity"), since if this weren't the case, $o$ would belong to $P\setminus\overrightarrow{ps}$. But then $\angle apo$ would have to be obtuse, since the complementary angle $\angle aps$ must be acute since it belongs to the right triangle $\triangle asp$. Hence by the Scalene inequality we would have $\overline{oa} > \overline{op}$, in contradiction to $o$ being the center of a circle that contains $a$ and $p$.
It is impossible that $o = p$, since this would contradict the fact that $o$ is the center of a circle that contains $a$ and $p$. So $o \in \overrightarrow{ps}\setminus\{p\}$.
Denote by $t$ the intersection of $cd$ and $P$. Let $q$ be any point on $P$ that is strictly farther from $p$ than both $o$ and $t$. Denote by $z$ the circle centered at $q$ that passes through $p$. We will show that $z$ is the desired circle, i.e. that the closed rectangle $abcd$ lies in the interior of $z$.
Since the interior of a circle is convex, it suffices to show that $a$, $b$, $c$, and $d$ all lie in the interior of $z$. By symmetry, it suffices to show that $a$ and $d$ lie in the interior of $z$, which is what we'll do.
To see that $a$ lies in the interior of $z$ it suffices to show that $\overline{qa} < \overline{qp}$. In fact, by the triangle inequality $\overline{qa} < \overline{qo} + \overline{oa}$, and since $a, p \in r$, $\overline{oa} = \overline{op}$.
To see that $d$ lies in the interior of $z$ it suffices to show that $\overline{qd} < \overline{qp}$. Since we have shown that $\overline{qa} < \overline{qp}$, it suffices to show that $\overline{qd} < \overline{qa}$. Since $\angle adq$ contains the right angle $\angle adc$, we conclude, by considering $\triangle adq$, that $\angle qda\ (= \angle adq) > \angle qad$, and therefore by the Scalene inequality $\overline{qa} > \overline{qd}$. $\square$
* Note that the proof makes no use of the assumption given in the original statement of the problem that $\overline{ab} \geq \overline{bc}$.
Bibliography
[V] Venema, Gerard A., Foundations of Geometry, 2nd edition, Pearson, 2012.